NamePros.Com - View Single Post - Randomingly choosing a quote from Database, PHP

Quote:

PHP Code:

    //get a random quote... using arrays.

 
//Be sure to connect to the db.

$query1 = "SELECT * FROM quotes";

$result1 = mysql_query($query1);

while ($row1 = mysql_fetch_array($result1))

{

    $id1 = $row1['id'];

    $array1[count($array1) + 1] = $id1;

    

    shuffle($array1);

}

shuffle($array1);

 
//ACCESS THE RANDOM NUMBER IS BY $array1[0]

//Now we have the id of the random quote.

 
$random = $array1[0];

 
$q1 = "SELECT * FROM quotes WHERE id='$random' LIMIT 1";

$r1 = mysql_query($q1);

 
while ($row1 = mysql_fetch_array($r1))

{

    $content = $row1['content'];

    $author = stripslashes($row1['author']);

}

 
//Then just echo $content and $author (the person who said it) when you need it.

?>

It may not be efficient but it works

it's returning this error aoubt how in

PHP Code:

    while ($row1 = mysql_fetch_array($r1))

$r1 is not a valid MySQL resource for mysql_fetch_array

I'm sure I did everything right (including connecting to the database). To prove it: when I replace

PHP Code:

    $q1 = "SELECT * FROM quotes WHERE id='$random' LIMIT 1";

with

PHP Code:

    $q1 = "SELECT * FROM quotes WHERE 1 LIMIT 1";

It returns with my first quote.

Just wondering...how do I fix the error? Thanks!